Differential Equations Question 3
Question 3 - 25 January - Shift 1
Let $y=y(x)$ be the solution curve of the differential equation $\frac{d y}{d x}=\frac{y}{x}(1+x y^{2}(1+\log _{e} x))$,
$x>0, y(1)=3$. Then $\frac{y^{2}(x)}{9}$ is equal to:
(1) $\frac{x^{2}}{5-2 x^{3}(2+\log _{e} x^{3})}$
(2) $\frac{x^{2}}{2 x^{3}(2+\log _{e} x^{3})-3}$
(3) $\frac{x^{2}}{3 x^{3}(1+\log _{e} x^{2})-2}$
(4) $\frac{x^{2}}{7-3 x^{3}(2+\log _{e} x^{2})}$
Show Answer
Answer: (1)
Solution:
Formula: Equations reducible to linear form (BERNOULI’S EQUATION)
$ \frac{d y}{d x}-\frac{y}{x}=y^{3}(1+\log _{e} x) $
$\frac{1}{y^{3}} \frac{dy}{dx}-\frac{1}{xy^{2}}=1+\log _{e} x$
Let $-\frac{1}{y^{2}}=t \Rightarrow \frac{2}{y^{3}} \frac{d y}{d x}=\frac{d t}{d x}$
$\therefore \frac{dt}{dx}+\frac{2 t}{x}=2(1+\log _{e} x)$
I.F. $=e^{\int \frac{2}{x} d x}=x^{2}$
$\frac{-x^{2}}{y^{2}}=\frac{2}{3}((1+\log _{e} x) x^{3}-\frac{x^{3}}{3})+C$
$y(1)=3$
$\frac{y^{2}}{9}=\frac{x^{2}}{5-2 x^{3}(2+\log _{e} x^{3})}$
